Problem: $\vec v = \left(-\dfrac34,-3\right)$ $4\vec v= ($
Solution: In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $4 \vec{v}$ : $\begin{aligned} {4}\vec v = {4} \cdot \left(-\dfrac34,-3\right) &= \left({4} \cdot \left(-\dfrac34\right), {4} \cdot (-3)\right) \\\\ &= (-3,-12) \end{aligned}$ The answer is $ (-3,-12) $.